HDU 5019：revenge of GCD【GCD】

原创于 2019-12-21 16:06:19 发布 · 175 阅读

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本文探讨了如何求解两个整数的最大公因数及其所有公因数，并通过编程实现，解决了求第k大的公因数的问题。特别讨论了当最大公因数为1时的特殊情况。

Discription
Now given two kinds of coins A and B,which satisfy that GCD(A,B)=1.Here you can assume that there are enough coins for both kinds.Please calculate the maximal value that you cannot pay and the total number that you cannot pay.

Input
The input will consist of a series of pairs of integers A and B, separated by a space, one pair of integers per line.

Output
For each pair of input integers A and B you should output the the maximal value that you cannot pay and the total number that you cannot pay, and with one line of output for each line in input.

Sample Input

2 3
3 4

Sample Output

1 1
5 3

题意
求两个数第k大的公因数
若因子数小于k则不存在，输出-1.

思路
求出两个数最大公约数，然后求出最大公因数的所有因数，存到一个vector里，从大到小排序，输出第k个。
需要特判最大公因数是1的情况。

AC代码

#include<bits/stdc++.h>
using namespace std;
typedef long long LL;

int t,f;
LL x,y,k;
vector<LL>v;
bool cmp(LL a,LL b)
{
    return a>b;
}

int main()
{
    scanf("%d",&t);
    while(t--)
    {
        v.clear();
        scanf("%lld %lld %lld",&x,&y,&k);
        LL tmp=__gcd(x,y);
        if(tmp==1&&k==1)
        {
            cout<<1<<endl;
            continue;
        }
        else if(tmp==1&&k>1)
        {
            cout<<-1<<endl;
            continue;
        }
        for(LL i=1; i<sqrt(tmp); i++)
            if(tmp%i==0)
            {
                v.push_back(i);
                v.push_back(tmp/i);
            }
        if((int)sqrt(tmp)*(int)sqrt(tmp)==tmp)
            v.push_back(sqrt(tmp));
        sort(v.begin(),v.end(),cmp);
        if(v.size()<k)
            cout<<-1<<endl;
        else
            cout<<v[k-1]<<endl;
    }
    return 0;
}